Problem: Factor the following expression: $x^2 + 14x + 48$
When we factor a polynomial, we are basically reversing this process of multiplying linear expressions together: $ \begin{eqnarray} (x + a)(x + b) &=& xx &+& xb + ax &+& ab \\ \\ &=& x^2 &+& {(a + b)}x &+& {ab} \end{eqnarray} $ $ \begin{eqnarray} \hphantom{(x + a)(x + b) }&\hphantom{=}&\hphantom{ xx }&\hphantom{+}&\hphantom{ (a + b)x }&\hphantom{+}& \\ &=& x^2 & +& {14}x& +& {48} \end{eqnarray} $ The coefficient on the $x$ term is $14$ and the constant term is $48$ , so to reverse the steps above, we need to find two numbers that add up to $14$ and multiply to $48$ You can try out different factors of $48$ to see if you can find two that satisfy both conditions. If you're stuck and can't think of any, you can also rewrite the conditions as a system of equations and try solving for $a$ and $b$ $ {a} + {b} = {14}$ $ {a} \times {b} = {48}$ The two numbers $6$ and $8$ satisfy both conditions: $ {6} + {8} = {14} $ $ {6} \times {8} = {48} $ So we can factor the expression as: $(x + {6})(x + {8})$